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The root system of the complex special linear Lie algebra ${\mathfrak{sl}}_6(\mathbb{C})$ is $\{e_i-e_j:1\le i\ne j\le 6\}$ in ${\sum }_i{x}_i=0$. Its simple roots $e_i-e_{i+1}$ give a five-node An Dynkin diagram. The defining fundamental representation has six weights $w_i=e_i-\frac{1}{6}{\sum }_j{e}_j$. The dual representation has weights $-w_i$, and the two-index antisymmetric representation has fifteen weights $w_i+w_j$, $i<j$, all with weight multiplicity one.

For a representation $R$ of $SU(N)$ with $N\ge 3$, choose the Hermitian quantum generator convention and normalize the cubic anomaly coefficient byHere the anticommutator symmetrizes the last two generators; cyclic property of the trace makes the whole tensor symmetric. This coefficient multiplies the local four-dimensional gauge anomaly from a chiral fermion in $R$. It adds under direct sums.

The conjugate representation has generators $-(T^a{)}^T$, so transposition and the three minus signs give $A(\overline{R})=-A(R)$. Hence a real representation or pseudoreal representation has zero cubic anomaly coefficient. For $SU(3)$, the two-index antisymmetric representation is the conjugate triplet and has coefficient $-1$. For $SU(2)$ the symmetric cubic invariant itself vanishes, so normalization to a nonzero fundamental cubic coefficient is unavailable; this statement concerns the local cubic anomaly and does not settle global anomalies.^[ref]

For the branching from B3 to A3, apply the supplied weight projection matrix to column vectors of Dynkin labels. It sendsIt is consistent with choosing the $A_3$ simple roots in the order ${\beta }_1=e_2-e_3$, ${\beta }_2=e_1-e_2$, ${\beta }_3=e_2+e_3$. Their Dynkin diagram is the chain ${\beta }_1-{\beta }_2-{\beta }_3$. Their coroot pairings are respectively $a_2,a_1,a_2+a_3$, explaining the projection without confusing labels and Euclidean coordinates.

The seven weights in b project asThe six nonzero weights form the weight system of the SU(4) two-index antisymmetric representation of highest weight $(0,1,0)$. The zero weight is the remaining singlet representation. ConsequentlyThe eight spinor weights project asThe four weights $(1,0,0),(-1,1,0),(0,-1,1),(0,0,-1)$ constitute the defining fundamental representation $4$. The other four are their negatives, forming the dual representation $\overline{4}$ of highest weight $(0,0,1)$. HenceFor the Young tableaux and tensors, let $U={\mathbb{C}}^4$ be the defining SU(4) representation. The singlet representation is ${\Lambda }^{0}U$, with the empty Young diagram and a scalar. The other three Young diagrams are columns of one, two and three boxes, representing $U$, ${\Lambda }^{2}U$ and ${\Lambda }^{3}U$:The square brackets indicate tensor index antisymmetrization, with indices from 1 to 4. The invariant Levi-Civita symbol identifies the three-index antisymmetric tensor with a covector, $w_{\ell }={ϵ}_{\ell ijk}{T}^{ijk}/3!$, explaining ${\Lambda }^{3}U\cong U^{\ast }$.

Write the defining space as $V=A\oplus B$, where $A\cong {\mathbb{C}}^2$ and $B\cong {\mathbb{C}}^3$. Use ordered pairs for the SU(2) and SU(3) factors, in that order. The generator $X_c$ in solution gives U(1) charges $3$ on $A$ and $-2$ on $B$, soTaking the conjugate representation reverses the U(1) charges. The SU(2) doublet is equivalent to its dual representation, while the SU(3) triplet becomes the antitriplet:

For the two-index antisymmetric representation, split the antisymmetric tensor $T^{IJ}=-T^{JI}$ into $T^{ab}$, $T^{a\mu }$ and $T^{\mu \nu }$, with $a,b=1,2$ and $\mu ,\nu =3,4,5$. The exterior square of a direct sum givesThe first term is an SU(2) singlet, the middle is $(2,3)$, and ${\Lambda }^{2}3\cong \overline{3}$ by contraction with the invariant alternating tensor. Adding the U(1) charges of the two indices givesThe dimensions $1+6+3=10$ check the branching of a representation. In the convention $Q_{\mathrm{em}}=T_3+Y$, weak hypercharge is $Y=X_c/6$.

For the embedding in solution, restriction retains the first four Dynkin label coordinates, since the simple coroots of $A_4$ are the first four of $A_5$. In a coordinate-free version of the same projection, let ${\eta }_i$ be the five defining weights of a representation of $A_4$. Restriction gives ${\epsilon }_i\mapsto {\eta }_i$ for $1\le i\le 5$, while ${\epsilon }_6\mapsto 0$.

For the fundamental representation, its first five weights of a representation therefore form a $5$ and the sixth is the zero weight of a representation of a trivial representation. Negating the weights of a representation gives the conjugate representation:For the two-index antisymmetric representation, split the fifteen weights of a representation into the ten with $i<j\le 5$ and the five with $j=6$. Their restricted values are ${\eta }_i+{\eta }_j$ and ${\eta }_i$, respectively. These are precisely the weight systems of $10$ and $5$. Negating again givesFor instance the highest $A_5$ weight of a representation $(0,1,0,0,0)$ restricts to $(0,1,0,0)$ for the $10$, while $(1,0,0,0,-1)$ restricts to $(1,0,0,0)$ for the $5$. The restricted weight systems are disjoint unions of multiplicity-one sets in these cases, and their dimensions check as $6=5+1$ and $15=10+5$.

Let $V={\mathbb{C}}^6$ be the defining fundamental representation of $SU(6)$. Its weights are $w_i=e_i-\frac{1}{6}{\sum }_{j=1}^6{e}_j$. Pairing with the simple coroots gives Dynkin labels $(w_i{)}_a={\delta }_{ia}-{\delta }_{i,a+1}$. Thus the complete list isSuccessive lowering operators subtract the corresponding rows of the Cartan matrix in a. Each weight multiplicity is one.

The complex-conjugate representation $\overline{6}\cong V^{\ast }$ is the dual representation and has the negatives of these weights. Its highest weight is $-w_6=(0,0,0,0,1)$, and its entire weight decomposition isEvery entry has weight multiplicity one.

The two-index antisymmetric representation is ${\Lambda }^{2}V=15$, of dimension $\left(\genfrac{}{}{0ex}{}{6}{2}\right)=15$ and highest weight $w_1+w_2=(0,1,0,0,0)$. A weight basis is $e_i\wedge e_j$ for $i<j$, carrying weight $w_i+w_j$. The fifteen sums, all distinct and all of weight multiplicity one, areThe exclusion of $i=j$ comes from the exterior square, where $e_i\wedge e_i=0$.

One generation of left-handed Standard Model fermions fits into $\overline{5}\oplus 10$ of an SU(5) grand unified theory. With hypercharge defined by $Q=T_3+Y$, their restriction of a representation to $SU(3{)}_c\times SU(2{)}_L\times U(1{)}_Y$ isHere $Q$ in the multiplet list denotes the left-handed quark doublet, and the superscript $c$ labels left-handed charge-conjugate fields of the corresponding right-handed fermions.

The results of c therefore giveThus the required matter representations are present. The extra singlet has Standard Model representation $(1,1{)}_0$ and can describe the left-handed charge-conjugate field of a right-handed neutrino. Its gauge singlet character permits a Majorana mass term; a Yukawa interaction with a lepton doublet and a suitable Higgs field can then supply neutrino masses. The branching alone does not fix those interactions or their scales.

The extra $5$ hasIf it describes Weyl fermions, these are additional coloured and electroweak fermions in the conjugate Standard Model representations to the ordinary $\overline{5}$. With an additional $\overline{5}$, they can form a vectorlike gauge spectrum and acquire invariant masses. Pairing them with the only ordinary $\overline{5}$ would also remove that generation's $d^c,L$ fields, so a realistic heavy pairing needs extra matter. If the $5$ instead describes a scalar field, it has the quantum numbers of the usual $SU(5)$ Higgs field: a weak doublet and a coloured triplet. A single fermionic $15$ does not turn its extra fermions into a scalar Higgs field merely by branching.

There is a further consistency condition if these are the entire chiral matter content of an $SU(6)$ gauge theory. Normalize the cubic anomaly coefficient by $A(6)=1$. The dual representation has $A(\overline{6})=-1$, and the two-index antisymmetric representation has $A(15)=6-4=2$. Thus one copy of each has nonzero gauge anomaly, $A_{\mathrm{total}}=1$. Their inclusion of the desired $SU(5)$ multiplets is therefore not, by itself, a complete consistent chiral unified theory.

For example, anomaly cancellation holds for $15\oplus 2\overline{6}$, whose restriction of a representation isThis leaves one ordinary generation, a potentially massive vectorlike gauge spectrum, and two possible singlet neutrinos. This anomaly-free matter choice is also used in explicit $SU(6)$ model constructions.^[ref] These interpretations require additional choices of matter statistics, spontaneous symmetry breaking, and interactions; the weight systems determine the multiplet content.

Use the full definition of the special unitary group: $A^{†}A=I$ and $\det A=1$. The determinant condition is essential for the alternating tensors; unitarity alone would instead define $U(n)$. Assume $n\ge 2$ when discussing proper invariant subspaces.

Let $V={\mathbb{C}}^n$ be the fundamental representation. A $(j,k)$ tensor belongs to $V^{\otimes j}\otimes (V^{\ast }{)}^{\otimes k}$. Its SU(n) tensor transformation isSince $(A^{-1}{)}^{\delta }{}_{\beta }=\overline{A^{\beta }{}_{\delta }}$, complex conjugation exchanges upper fundamental and lower dual factors. Reordering the two index groups shows that $\overline{T}$ is a $(k,j)$ tensor.

The invariant tensors follow directly. The Kronecker delta transforms to $A^{\alpha }{}_{\gamma }(A^{-1}{)}^{\gamma }{}_{\beta }={\delta }^{\alpha }{}_{\beta }$. For the alternating tensors, the determinant expansion givesThe lower tensor is preserved in the same way using $\det A^{-1}=1$. Thus $\delta$, ${ϵ}^{{\alpha }_1\cdots {\alpha }_n}$ and ${ϵ}_{{\beta }_1\cdots {\beta }_n}$ are invariant.

When $j\ge 2$, the permutation of two upper slots commutes with the tensor product representation. Its $+1$ and $-1$ eigenspaces are nonzero proper invariant subspaces. The same argument uses two lower slots when $k\ge 2$. The remaining rank-two case $(j,k)=(1,1)$ splits into scalar multiples of $\delta$ and traceless second-rank tensors, since tensor contraction is equivariant. This covers every case except the fundamental, dual fundamental and scalar cases specified in the question. Those three are irreducible: the special unitary group acts transitively on unit vectors in its fundamental space, so a nonzero invariant subspace must be the full space, and the dual has the same property.

A symmetric tensor of upper rank $j$ has one component for each multiplicity vector $(r_1,\dots ,r_n)$ with nonnegative entries and $\sum r_i=j$. Equivalently it is a homogeneous degree-$j$ polynomial in $n$ variables. Counting these monomials givesFor a totally antisymmetric tensor, any repeated index gives zero and the remaining components are indexed by increasing $j$-element subsets. HenceThese are the dimensions of the symmetric power and exterior power, respectively.

For SU(4), the two-index antisymmetric representation has complex dimension six. To establish a six-dimensional real representation, one must exhibit a real structure; simply counting six complex components would give twelve real components. Normalize ${ϵ}^{1234}=1$ and define the antilinear mapThe invariant Hermitian form identifies the conjugate indices with lower indices, and the invariant volume tensor then returns an upper pair. Thus $\mathcal{J}$ commutes with SU(4). The identitygives ${\mathcal{J}}^2=1$. Its fixed space hasThree arbitrary complex entries specify all the others, giving exactly six real degrees of freedom. The real structure of the SU4 exterior square therefore provesThe ordinary Hermitian norm restricts to an invariant positive real inner product on this fixed space.

Irreducible representations classify states because an exact symmetry commutes with the Hamiltonian, preserving energy eigenspaces. Irreducible multiplets are the smallest sets closed under all symmetry transformations; their labels and Casimir operators provide quantum numbers. Approximate flavour symmetry gives approximate multiplets and mass relations, with splittings caused by its breaking. Spin and internal quantum numbers are distinct labels.

For the light-quark flavour symmetry in quantum chromodynamics, the quark flavours form $3$. Their Triple tensor decomposition for the defining sl3 representation isThis algebraic decomposition alone does not determine the allowed ground-state baryons. The three-quark colour singlet is antisymmetric. The no-orbital-excitation ground state has a symmetric spatial wavefunction, so the Pauli exclusion principle requires the combined spin-flavour wavefunction to be symmetric. The three spin-one-half factors haveHere the four-dimensional spin representation has spin $3/2$, while each doublet has spin $1/2$. Symmetric decuplet flavour pairs with symmetric spin $3/2$. The two flavour-octet copies and two spin-doublet copies carry the two-dimensional standard permutation multiplicity space; its tensor square contains one symmetric singlet. Their invariant pairing therefore supplies one octet with spin $1/2$, not two independent ground-state octets. Antisymmetric flavour would need a completely antisymmetric three-quark spin state, which does not exist.

Equivalently the symmetric spin-flavour SU6 representation hasThus the Pauli constraint on three-quark flavour multiplets gives the concise spectrum classificationThese are the baryon octet and baryon decuplet. The three-quark flavour singlet requires a different spatial permutation symmetry and belongs to excited states, outside the symmetric ground-state approximation used here. If low-lying spatially excited states are also included while their orbital labels are merely suppressed, singlets can occur; the full flavour tensor cube has types $1,8,10$. The absence of the singlet from the unexcited ground multiplet is a permutation-symmetry constraint, not a prohibition by $SU(3)$ alone. Light-quark mass differences make the flavour symmetry approximate rather than exact.

Normalize the cubic anomaly coefficient of the $SU(N)$ fundamental to $+1$. The two-index antisymmetric representation has anomaly coefficient $N-4$, while an antifundamental contributes $-1$. Cancellation of the $SU(N{)}^3$ gauge anomaly requires

The electric quarks give $2N_c$ copies of the $SU(2N_f)$ fundamental, soIn the magnetic theory the $2{\tilde{N}}_c$ color components of $q$ transform as flavor antifundamentals and contribute $-2{\tilde{N}}_c$. The singlet $M^{ij}$ is the two-index antisymmetric representation of $SU(2N_f)$, whose cubic anomaly coefficient is $2N_f-4$. HenceThis is a direct anomaly check of Seiberg duality.

The four-by-four unitary matrices of determinant one form this special unitary group. Its two-index antisymmetric representation has complex dimension six and the real structure of the SU4 exterior square realizes it on a six-dimensional real space.